Answer:
The concentration of KOH is 0.186 M
Explanation:
First things first, we need too write out the balanced equation between HBr and KOH.
This is given as;
KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)
From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.
We use the acid base formular in calculating unknown concentrations. This is given as;
where;
Ca = Concentration of acid
Va = Volume of acid
Cb = Concentration of base
Vb = Volume of base
na = Number of moles of acid
nb = Number of moles of base
KOH is the base and HBr is acid.
Hence;
Ca = 0.225
Va = 35
Cb = ?
Vb = 42.3
na = 1
nb = 1
Making Cb subject of formular we have;
Cb = (0.225 * 35 * 1) / (42.3 * 1)
Cb = 0.186 M
We know that:
mass = density x volume
The volume of the total mixture is 0.1 m³
Let the first liquid be A and the second be B
Mass Total = Mass A + Mass B
800 x 0.1 = 1500Va + 500(0.1 - Va)
30 = 1000Va
Va = 0.03 m³
Vb = 0.1 - 0.03 = 0.07 m³
1 mol = 6.022 x 10²³ atoms
In order to find how many atoms, dimly multiply the amount of moles you have by 6.022 x 10²³ or Avogadro's number.
So you have 1.75 mol CHC1₃ x (6.022x10²³) = 1.05385 x 10²⁴ atoms of CHCl₃
But now you have to round because of the rules of significant figures so you get 1.05 x 10²⁴ atoms of CHCl₃
There are 13 atoms in the product
<h3>Further explanation</h3>
Given
Reaction
H2SO4 + 2KOH --> K2SO4 + 2H2O
Required
The number of atoms
Solution
In a chemical equation, there are reactants on the left and products on the right
Reactants : H2SO4 + 2KOH
Products : K2SO4 + 2H2O
The number of atoms is determined by their reaction coefficient and the subscript of the atoms in the compound
K2SO4 (coefficient = 1) :
K = 2 atoms
S = 1 atom
O = 4 atoms
Total atoms = 7 atoms
2H2O(coefficient = 2) :
H = 2 x 2 = 4 atoms
O = 2 x 1 = 2 atoms
Total atoms = 6 atoms
Total = 13 atoms
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